If the thermal conductivity of a material is 0.030 W/m·K and the heat load is 500 W, what is the thickness of the styrofoam insulation needed?
Understand the Problem
The question is asking to determine the thickness of styrofoam insulation required, given the thermal conductivity and heat load. To solve this, we will use the formula related to heat transfer through a material, which connects thermal conductivity, heat load, and thickness.
Answer
The thickness $d$ can be calculated using the formula $d = \frac{k \cdot A \cdot \Delta T}{Q}$.
Answer for screen readers
The thickness of the styrofoam insulation required is given by
$$ d = \frac{k \cdot A \cdot \Delta T}{Q} $$
after substituting the specific values.
Steps to Solve
-
Identify the formula for heat transfer
The heat transfer through a material can be expressed using the formula:
$$ Q = \frac{k \cdot A \cdot \Delta T}{d} $$
Where:
- $Q$ is the heat load (in Watts)
- $k$ is the thermal conductivity of the material (in W/m·K)
- $A$ is the area of the material (in m²)
- $\Delta T$ is the temperature difference across the material (in K)
- $d$ is the thickness of the material (in meters)
-
Rearranging for thickness
To find the thickness ($d$), we will rearrange the formula:
$$ d = \frac{k \cdot A \cdot \Delta T}{Q} $$ -
Substituting values into the formula
Next, we will plug in the given values for $k$, $A$, $\Delta T$, and $Q$ into the rearranged formula and solve for $d$. Ensure that all units are consistent. -
Calculating the thickness
Perform the calculations step-by-step to determine the value of the thickness $d$.
The thickness of the styrofoam insulation required is given by
$$ d = \frac{k \cdot A \cdot \Delta T}{Q} $$
after substituting the specific values.
More Information
This calculation is important for determining how much insulation is needed to improve energy efficiency in buildings. Styrofoam is commonly used for its low thermal conductivity.
Tips
- Forgetting to convert units properly (e.g., m² to cm², or W to kW).
- Using the wrong values for thermal conductivity or heat load.
- Misinterpreting the formula and incorrectly rearranging it.
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